# Why can quantum walks not approach a stationary distribution

In Child's notes on quantum walks, he claims (section 16.6) "Since a quantum walk is a unitary process, we should not expect it to approach a limiting quantum state, no matter how long we wait."

But why should this be true? I understand that quantum mechanical processes must be time-reversible, but so long as the distribution is only stationary in the limit, it seems fine to allow quantum walks to merely approach a limiting state.

• The fact that quantum walks reach no stationary distribution holds true only for closed systems. Dec 15, 2022 at 15:12

Suppose for contradiction that there is some limiting state $$|f\rangle$$ that an initial state $$|s_1\rangle \neq |f\rangle$$ approaches as a unitary operation $$U$$ is repeatedly applied. So there is some series of states $$|s_1\rangle$$, $$|s_2\rangle$$, etc where $$|s_{k+1}\rangle = U |s_k\rangle$$ and $$\lim_{k \rightarrow \infty} |s_k\rangle = |f\rangle$$.

Here's the problem. Unitary operations can't change relative dot products. It's always the case that $$|\langle a | b \rangle| = |\langle Ua | Ub \rangle|$$. Consider the overlap between the state of the iteration starting from $$|s_1\rangle$$, and what you would have gotten instead by starting from $$|s_2\rangle$$, as you repeatedly apply $$U$$. The initial overlap between the current state of these processes is some number less than one. But by assumption of the existence of $$|f\rangle$$ the action of iterating $$U$$ needs to send them both towards $$|f\rangle$$. The action needs to increase their overlap in a way that limits to 1. Iterating $$U$$ needs to increase their overlap, but unitary operations can't increase overlap. Contradiction. Therefore there is not actually a limiting state $$|f\rangle$$.

• Great answer, this makes perfect sense. Thanks! Dec 15, 2022 at 7:00

That's an interesting point and I suppose reversibility, interpreted as bijectivity, is not enough.

But unitarity implies a lot more. More specifically, unitary operators preserve distance. So we can prove a short impossibility result on limits existing, i.e., $$U^n\vert x\rangle$$ can never converge unless $$\vert x\rangle$$ is already an eigenvector of eigenvalue $$1$$.

Suppose for a contradiction that $$U^nx$$ converges to some state $$y$$ (I won't use kets just to make notation easier). Let $$\epsilon > 0$$. Pick whatever $$N_1$$ is large enough such that $$\Vert U^N x - y\Vert < \epsilon$$ for all $$N\geq N_1$$. But notice that $$U^{N_1}$$ is also unitary, so we can argue that

$$\Vert U^{N_1}x - U^{2N_1}x\Vert = \Vert U^{N_1}(x - U^{N_1}x)\Vert = \Vert x - U^{N_1}x\Vert$$

and then use the triangle inequality

$$\Vert U^{N_1}x - U^{2N_1}x\Vert \leq \Vert U^{N_1}x - y\Vert + \Vert y - U^{2N_1}x\Vert < 2\epsilon$$

but then we can further argue that

$$\Vert x - y\Vert \leq \Vert x - U^{N_1}x\Vert + \Vert U^{N_1}x - y\Vert < 3\epsilon$$ by combining the equation and inequality above. Since $$\epsilon$$ was arbitrary, this implies $$x=y$$.